Question

$$\frac{dy}{dx}=2e^{2x}{y}^2,y\left(0\right)=-1$$

Answer 1

$$\frac{dy}{dx}=2e^{2x}{y}^2,y\left(0\right)=-1$$
$$\Rightarrow \frac{1}{y^2}dy=2e^{2x}dx$$
Integrating both sides, we get
$$\int \frac{1}{y^2}dy=2\int e^{2x}dx$$
$$\Rightarrow \frac{-1}{y}=e^{2x}+C.....(1)$$
We know that at x = 0, y = - 1 . 
Substituting the values of x and y in (1), we get
$$1=1+C$$
$$\Rightarrow C=0$$
Substituting the value of C in (1), we get
$$-\frac{1}{y}=e^{2x}$$
$$\Rightarrow y=-e^{-2x}$$
$$\text{ Hence, }y=-e^{-2x}\text{ is the required solution }.$$